MHB Is There a Power of 3 That Ends in 001?

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A power of 3 that ends in 001 can be proven to exist by considering the integers 3^1 through 3^1001 and applying the pigeonhole principle, which indicates that two distinct powers must yield the same remainder when divided by 1000. This leads to the conclusion that there are distinct integers i and j such that 3^i is congruent to 3^j modulo 1000. Consequently, it follows that 3^(j-1) is congruent to 1 modulo 1000, confirming the existence of such a power. The discussion also raises the question of which odd numbers can replace 001, suggesting that the count of these numbers under 1000 is related to the divisor of φ(1000). The exploration of these odd replacements remains open for further investigation.
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Prove that there exists a power of 3 that ends in 001.
 
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mathmaniac said:
Prove that there exists a power of 3 that ends in 001.
consider the integers:
$3^1,3^2,\ldots,3^{1001}$

Some two distinct integers of these must leave the same remainder mod $1000$.
So there exist distinct $i$ and $j$ such that $3^i\equiv 3^j\pmod{1000}$.
WLOG $i<j$. Thus $3^i(3^{j-i}-1)\equiv 0\pmod{1000}$
Thus $3^{j-1}\equiv 1\pmod{1000}$ and we are done.
 
Good work!So quick!
Now what odd numbers can we replace for 001?
 
Last edited:
mathmaniac said:
Good work!So quick!
Now what odd numbers can we replace for 001?
I don't know if there's an analytic way to enumerate all such numbers. But it can be shown then the number of such numbers which are less then 1000 divides $\phi(1000)$.
 

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